Genetics
- Monohybrid cross (Aa × Aa)
- genotypes 1 : 2 : 1, phenotypes 3 : 1
- Dihybrid cross
- phenotypes 9 : 3 : 3 : 1
- Hardy–Weinberg
- p + q = 1, p² + 2pq + q² = 1
- Independent genes
- multiply the probabilities
Biology: Genetics and Cell Biology: cram pack
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
1Formulas and methods
2Practice problems
1.GeneticsMedium
Tall (A) is dominant over short (a), and round seeds (B) over wrinkled (b). The genes are on different chromosomes. For the cross AaBb × aaBb, what is the probability P that an offspring is tall, wrinkled seeds?
AaBb × aaBb
P =
2.GeneticsHard
In a population of 500 in Hardy–Weinberg equilibrium, 1% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 500q² = 0.01
N =
3.Cell biologyMedium
Which motor protein generally carries cargo along microtubules towards the minus end, near the cell centre?
- AKinesin
- BMyosin
- CDynein
- DActin
4.Cell biologyHard
In the intrinsic (mitochondrial) pathway of apoptosis, which event leads to caspase activation?
- AFusion of lysosomes with the nucleus
- BBinding of a death ligand to the Fas receptor on the cell surface
- CEntry of glucose into the mitochondria
- DRelease of cytochrome c from mitochondria into the cytosol
3Solutions
1.Genetics
Answer: P = 0.125
Mendelian genetics: each parent passes on one allele of each gene, chosen at random. A Punnett square lists every combination of gametes, each equally likely. A dominant allele (capital letter) shows whenever it is present; a recessive one only in the homozygous form. Genes on different chromosomes are inherited independently, so their probabilities multiply.
Each gene on its own
P(tall) = 0.5
P(wrinkled seeds) = 0.25
Multiply
P = 0.5·0.25 = 0.125
Answer
P = 0.125
2.Genetics
Answer: N = 405
Hardy–Weinberg: in a large, randomly mating population with no selection, the allele frequencies p (dominant) and q (recessive), p + q = 1, give genotype frequencies p², 2pq and q².
Allele frequencies
q = √0.01 = 0.1
p = 0.9
Homozygous dominant
p² = 0.81
N = 500·0.81 = 405
Answer
N = 405
3.Cell biology
Answer: C · Dynein
Why this is the answer
Kinesins mostly move towards the plus end (out to the periphery) and dyneins towards the minus end (towards the centrosome). Myosin moves along actin.
Answer
C · Dynein
4.Cell biology
Answer: D · Release of cytochrome c from mitochondria into the cytosol
Why this is the answer
Cell stress lets pro-apoptotic Bcl-2 family proteins release cytochrome c. With Apaf-1 it forms the apoptosome, which activates caspase-9. Fas binding starts the extrinsic pathway.
Answer
D · Release of cytochrome c from mitochondria into the cytosol
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